Q.
Two equal metal balls are charged to 10 and -20 units of electricity. Then they are brought in contact with each other and then again separated to the original distance. The ratio of magnitudes of the force between the two balls before and after contact is
A) 8:1
B) 1:8
C) 2:1
D) 1:2
Explanation
(a) Given,$\mathcal{Q}_{1} = 10unit,\mathcal{Q}_{2} = - 20\ unit$
After contact charges on both become same,$ie,$
$\ \ \ \ \ \ \ \ \ \ \mathcal{Q}^{'} = \frac{10 - 20}{2} = - 5\ units$
$\left| F_{1} \right| = \frac{1}{4\pi\varepsilon_{0}} \times \frac{10 \times 20}{r^{2}}$
and $\left| F_{2} \right| = \frac{1}{4\pi\varepsilon_{0}} \times \frac{5 \times 5}{r^{2}}$
Hence, $\frac{{|F}_{1|}}{\left| F_{2} \right|} = \frac{8}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.